Are compact subsets of the topologist's sine circle, equipped with the Hausdorff distance, path connected? I think that the topologist's sine circle $C$ is an example of a path connected space such that $\mathcal K(C)$ is not path connected, where $\mathcal K(C)$ denotes compact subsets of $C$ equipped with the Hausdorff distance. Let $K_1$ be a point, and let $K_2$ be $C \cap H$, where $H$ is the closed upper half plane. I suspect that there is no continuous path from $K_1$ to $K_2$ (try imagining what such a thing would look like), but don't know how to approach showing this. I don't know how to wrangle anything useful out of the notion of a continuous path in $\mathcal K(C)$.
It's not hard to show that if $X$ is connected, so is $\mathcal K(X)$. I read the corresponding question about path connectedness a while back in C.C. Pugh's Real Mathematical Analysis.
The topologist's sine circle is this thing: 

When I say the upper half plane, I'm imagining the topologists sine curve part being cut in half by the x-axis. 
 A: $\mathcal{K}(C)$ is path connected.
Let $d$ be the usual Euclidean metric on $C$ and let $d_H$ denote the corresponding Hausdorff distance on $\mathcal{K}(C)$.  Let $K_\epsilon = \{x : d(x,K) \le \epsilon\}$ denote the closed $\epsilon$-fattening of a compact set $K$, and recall that if $K' \subset K_\epsilon$ and $K \subset K'_\epsilon$ then $d_H(K, K') \le \epsilon$.
Define $f : (0,1] \to C$ by $f(t) = (t, \sin(1/t))$ for $0 < t \le 1/2$, and for $1/2 \le t \le 1$ let it cover the rest of $C$ in any continuous manner (it won't be injective but that's okay).  Let $x_1 = f(1)$.
Let $K \subset C$ be a nonempty compact set.  
First, let us show there is a continuous path from $K$ to $K \cup \{x_1\}$. Choose any $t_0$ with $f(t_0) \in K$, and define $\sigma(t) = K \cup \{f(t)\}$ for $t_0 \le t \le 1$.  Clearly $\sigma(t_0) = K$ and $\sigma(1) = K \cup \{x_1\}$, and the continuity of $\sigma$ follows immediately from continuity of $f$.
Now let $\gamma(t) = K \cup f([t,1])$ for $0 < t \le 1$, and set $\gamma(0) = C$.  Clearly $\gamma(t)$ is compact for each $t$, and $\gamma(1) = K \cup \{x_1\}$.  I claim $\gamma$ is continuous.
To show it is continuous at 0, fix $\epsilon >0$ and assume without loss of generality that $\epsilon < 1/2$.  I claim that if $t < \epsilon$ we have $\gamma(t)_\epsilon = C$ and thus $d_H(\gamma(t), \gamma(0)) = d_H(\gamma(t), C) \le \epsilon$.  For if $p \in C \setminus \gamma(t)$ then $p = f(s) = (s, \sin(1/s))$ for some $0 < s < t < \epsilon$.  Now the point $q = (0, \sin(1/s))$ is contained in $f([1/2,1])$ by assumption, hence contained in $\gamma(t)$, and $d(p,q) = s < \epsilon$, so $p \in \gamma(t)_\epsilon$.
Now fix any $t_0 > 0$; we will show $\gamma$ is continuous at $t_0$.  By continuity of $f$ we can find $\delta$ such that if $|t-t_0| < \delta$ then $d(f(t), f(t_0)) < \epsilon$.  Fix such a $t$ and suppose that $t \ge t_0$.  We have $\gamma(t) \subset \gamma(t_0) \subset \gamma(t_0)_{\epsilon}$ by construction. If $p \in \gamma(t_0) \setminus \gamma(t)$ then $p = f(s)$ for some $s \in [t_0, t]$.  This means $|s-t| < \delta$, so letting $q = f(t)$, we have $d(p,q) = d(f(s), f(t)) < \epsilon$.  Hence $q \in \gamma(t)_\epsilon$ so $\gamma(t_0) \subset \gamma(t)_\epsilon$, and we have shown $d_H(\gamma(t), \gamma(t_0)) < \epsilon$.  By symmetry, the same holds if $t \le t_0$.
Here is an animation to help visualize this path.
https://youtu.be/WzjJFCj2whM
